combinatorics - Prove by a combinatorial argument that ${n choose r}{r choose s}={n choose s} {n-s choose r-s} $

Prove by a combinatorial argument that ${n \choose r}{r \choose s}={n \choose s} {n-s \choose r-s} $

Is a little hard for me solve this problem.

I see we need to use the multiplication principle. But is a little hard to me finding the idea for prove this...